[Linear Algebra] Inverse Matrix

Jason Lee·2022년 8월 5일
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Linear Algebra

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Inverse Matrix

  • Definition : For a square matrix ARn×nA \in \mathbb{R}^{n \times n}, its inverse matrix A1A^{-1} is defined such that A1A=AA1=InA^{-1}A = A A^{-1} = I_n

Solving Linear System via Inverse Matrix

  • We can now solve Ax=bA \textbf{x} = \textbf{b} as follows

Ax=bA \textbf{x} = \textbf{b}
A1Ax=A1bA^{-1} A \textbf{x} = A^{-1} \textbf{b}
Inx=A1bI_n \textbf{x} = A^{-1} \textbf{b}
x=A1b\textbf{x} = A^{-1} \textbf{b}

e.g.

Ax=bA \textbf{x} = \textbf{b}

where A=[221310131]A = \begin{bmatrix} 2 & 2 & 1 \\ 3 & 1 & 0 \\ 1 & 3 & 1 \\ \end{bmatrix}, x=[x1x2x3]\textbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, b=[61418]\textbf{b} = \begin{bmatrix} 6 \\ 14 \\ 18 \end{bmatrix}

A1=[0.250.250.250.750.250.75211]A^{-1} = \begin{bmatrix} 0.25 & 0.25 & -0.25 \\ -0.75 & 0.25 & 0.75 \\ 2 & -1 & -1 \\ \end{bmatrix}

x=A1b=[0.250.250.250.750.250.75211][61418]=[0.512.55]\textbf{x} = A^{-1} \textbf{b} = \begin{bmatrix} 0.25 & 0.25 & -0.25 \\ -0.75 & 0.25 & 0.75 \\ 2 & -1 & -1 \\ \end{bmatrix}\begin{bmatrix} 6 \\ 14 \\ 18 \end{bmatrix} = \begin{bmatrix} 0.5 \\ 12.5 \\ -5 \end{bmatrix}

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